Explanation
L=limx→π/62sin2x+sinx−12sin2x−3sinx−1L=2sin2(π6)+sin(π6)−12sin2(π6)−3sin(π6)−1L=2.14+12−12.14−3.12−1=0−2=0
So option D is correct
The function f:R/0→R given by f(x)=1x−2e2x−1 can be made continuous at x=0 bydefining f(0) as
limx→0logtan2x(tan22x)
Let,
y=logtan2xtan22x=logtan22x=(tan2x)y=2logtan2x=2ylogtanx⇒y=logtan2xlogtanx⇒limx→0y=limx→0logtan2xlogtanx
Using L-Hospital Rule
⇒limx→0y=limx→01tan2x×sec22x×21tanxsec2x⇒limx→0y=limx→02sec22xtan2x×tanxsec2x⇒limx→0y=limx→02tanxtan2x(limx→0secx=4)
⇒limx→0y=limx→02tanxx×xtan2x2x×2x⇒limx→0y=limx→01(limx→0tanxx=1)⇒limx→0logtan2x(tan22x)=1
We have,
limx→0(e4x−1x)
This is 00 form.
So, by using L-Hospital rule
limx→0(e4x×4−01)
limx→0(4e4x)
=4×e0
=4×1
=4
Hence, this is the answer.
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